Towards a contradiction suppose that \(\Lambda _x\)  is infinite for some \(x\in \Omega \) . Let \(J^{\prime }\)  be a subinterval of \(\Lambda _x\)  where \(\preccurlyeq _x\)  is either \(\le \) , \(\ge \)  or \(\le \cup \ge \) . If \(\preccurlyeq _x\)  equals \(\le \cup \ge \)  then, for every \(t\in J^{\prime }\) , the sets \(U(x,t)\)  and \(L(x,t)\)  are empty, contradicting that \(\dim (L(x,t) \cup U(x,t))=\dim \Omega \) . Suppose that \(\preccurlyeq _x\)  is either \(\le \)  or \(\ge \) . Note that, for any distinct \(s,t\in J^{\prime }\) , the sets \(L(x,s)\) , \(U(x,s)\) , \(L(x,t)\)  and \(U(x,t)\)  are pairwise disjoint. Using the fact that \(\dim (L(x,t) \cup U(x,t))=\dim \Omega \)  for every \(t\in J^{\prime }\) , and applying the Fiber Lemma for o-minimal dimension [1]}, we derive that
\(\dim \left(\bigcup _{t\in J^{\prime }} (L(x,t) \cup U(x,t))\right) > \dim (\Omega ),\) 
